It worth mention that the delta function has a special meaning in quantum mechanics.
However, we assume that if the square of the delta function is well defined, the square root of the delta function itself can be found.
Can we trust the meaning of this statement?
The delta function is an operator, whose relevance isn't special to quantum mechanics.
Its square root can be found just as for any operator.
What is special about the delta function is that it's a singular operator, which means that a limiting procedure is required to define it.
But no one goes about calculating the operator, whose square is the delta function, unless there's a particular calculation to be done, in which case one sets up the limiting procedure and does it.
The square of the delta function is a mathematical concept that is not well-defined as a distribution. A distribution is a generalization of a function that can act on other functions, such as integration. The delta function, denoted by δ(x), is a distribution that is zero everywhere except at the origin, where it is infinite, and has the property that ∫−∞∞δ(x)f(x)dx=f(0) for any smooth function f(x). However, the product of two distributions is not always defined, and in particular, the square of the delta function δ(x)2 does not exist as a distribution. This is because there is no consistent way to assign a value to ∫−∞∞δ(x)2f(x)dx for any smooth function f(x). One way to see this is to consider a sequence of smooth functions S(x,t) that converge weakly to δ(x) as t goes to zero. For example, one such sequence is S(x,t)=14πte−x2/4t, which is a Gaussian function with mean zero and variance 2t. As t approaches zero, S(x,t) becomes more and more concentrated near the origin, and its integral over the whole real line remains equal to one. However, if we try to define δ(x)2 as the weak limit of S(x,t)2 as t goes to zero, we run into a problem. The integral of S(x,t)2 over the whole real line is equal to 12πt−−√, which goes to infinity as t goes to zero. Therefore, there is no finite value that we can assign to ∫−∞∞δ(x)2f(x)dx by taking the limit of ∫−∞∞S(x,t)2f(x)dx as t goes to zero. This shows that δ(x)2 is not well-defined as a distribution.
To answer that question you can go to one of the delta function representations (before the limit is taken)
for example there is a Lorenz representation, take the square root, and then again take the limit and see what results, if anything.
There is also an exponential one, and also an integral form in Fourier theory.
Lyudimil is probably correct, you can have better luck with derivatives of the delta function, which do exist.
Use integration by parts.
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